Problems Based on Permutations and Combinations

How many different ways can the letters of the word ‘LEADING’ be arranged so that the vowels always come together?

• The word ‘LEADING’ has seven different letters.
• When the vowels EAI are always together, they can be treated as one letter.
• Then, we have to arrange the letters LNDG(EAI).
• Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.
• The vowels (EAI) can be arranged in 3! = 6 ways.

Therefore, the required number of ways = (120 x 6) = 720.

How many different ways can the letters of the word ‘CORPORATION’ be arranged so that the vowels always come together?

• In the word ‘CORPORATION’, we treat the vowels OOAIO as one letter.

• Thus, we have CRPRTN (OOAIO).
• This has 7 (6 + 1) letters, of which R occurs two times, and the rest are different.
• Number of ways arranging these letters = 7!/2! = 2520.
• Now, 5 vowels in which O occurs 3 times, and the rest are different, can be arranged in 5!/3! = 20 ways.

Therefore, the required number of ways = (2520 x 20) = 50400.

How many different ways can the letters of the word ‘MATHEMATICS’ be arranged so that the vowels always come together?

• We treat the vowels AEAI as one letter in the word ‘MATHEMATICS’.
• Thus, we have MTHMTCS (AEAI).
• Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice, and the rest are different.
• Number of ways of arranging these letters = 8!/(2!)(2!) = 10080.
• Now, AEAI has 4 letters in which A occurs 2 times, and the rest are different.
• Number of ways of arranging these letters = 4!/2! = 12.

Therefore, the required number of words = (10080 x 12) = 120960.

In how many ways can 30 identical apples be divided among 10 boys?

Here, the apples are identical, and boys are distinct. Number of ways in which 30 identical apples can be divided among 10 boys = Number of ways in which 30 identical balls can be distributed into 10 distinct boxes This problem can be solved using the formula:
No of ways=^(k+n-1)C_(n-1)
Here n = 10, k = 30.
Hence, the required number of ways
= ^(k+n-1)C_(n-1) = ³⁹C₉ =39!/9!30!

A company has 11 software engineers and seven civil engineers. How many ways can they be seated in a row so that all the civil engineers are always together?

• All 7 civil engineers are always together. Hence, group all the 7 civil engineers and consider them a single civil engineer. Hence, we can take the total number of engineers as 12.
• These 12 engineers can be arranged in 12! ways …(A)
• We had grouped seven civil engineers. These 7 civil engineers can be arranged among themselves in 7! ways …(B)

From (A) and (B), the required number of ways = 12! × 7!

There are three places P, Q, and R such that 3 roads connect P and Q, and four roads connect Q, and R. How many ways can one travel from P to R?

The number of ways in which one can travel from P to R = 3 × 4 = 12

What is the sum of all 4 digit numbers formed using the digits 2, 3,4, and 5 without repetition?

Sum of all 4 digit numbers formed using the digits  2,3,4,5  without repetition = (4−1)! (2+3+4+5)(1111) = 3!×14×1111 = 6×14×1111 = 93324

In how many ways can 10 books be arranged on a shelf such that a particular pair of books should always be together?

• We have a total of 10 books.
• Given that a particular pair of books should always be together. Hence, just tie these two books together and consider their single books.
• Hence we can take the total number of books like 9. These nine books can be arranged in ⁹P₉ = 9! ways. We had tied two books together. These books can be arranged among themselves in ²P₂ = 2! ways.

Hence, the required number of ways = 9! × 2!

How many ways can 11 persons be arranged in a row so that 3 particular persons should always be together?

• Given that three particular persons should always be together. Hence, just group these three persons and consider them single people.
• Therefore we can take the total number of persons as 9. These 9 persons can be arranged in 9! ways.
• We had grouped three people. These three persons can be arranged among themselves in 3! ways.

Hence, the required number of ways = 9! × 3!.

Five balls need to be placed in three boxes. Each box can hold all five balls. How many ways can the balls be placed in the boxes so that no box remains empty if all balls and boxes are identical?

Since no box can be left empty, there can be only two cases. Case A: 1,1,3 (i.e., three balls are put in 1 box, and 1 ball is put in each of the remaining 2 boxes). Total number of ways in which this can be done = 1 (as both boxes and balls are identical) Case B: 1,2,2 (i.e., two balls are put in each of the two boxes, and 1 ball is put in the remaining 1 box). The total number of ways in which this can be done = 1 (as both boxes and balls are identical). From (A) and (B), Required number of ways = 1+1=6

Five balls need to be placed in three boxes. Each box can hold all five balls. How many ways can the balls be placed in the boxes so that no box remains empty? What if all balls and boxes are identical, but the boxes are placed in a row?

Since no box can be left empty, there can be only two cases. Case A: 1,1,3  (i.e., 3 balls are put in 1 box, and 1 ball is put in each of the remaining 2 boxes). A box (in which 3 balls are put) can be selected 3C1 ways. Now, the three balls can be selected only in 1 way (as all the balls are identical). The remaining 2 balls can be arranged in only 1 way (as all the balls are identical).
Hence, the total number of ways = ³C₁ = 3      …(A)
Case B:
1,2,2
(i.e., two balls are put in each of the two boxes, and 1 ball is put in the remaining 1 box).
The two boxes (in each of them, two balls are put) can be selected in 3C2 ways.

• Now, two balls for the first selected box can be selected only in 1 way(as all the balls are identical).
• Two balls for the second selected box can be selected in only 1 way(as all the balls are identical).
• The remaining 1 ball can be placed only in 1 way.
• Hence, the total number of ways = ³C₂ = 3  …(B)

From (A) and (B), the required number of ways = 3 + 3 = 6.